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Use an arbitrary mass, 100 g is an easy number to work with.
60% of 100 g is 60 g, there are two A's. Each A is 30 g
40 g is B, and there is only one, so B is 40 g.
<span>A<span>B2</span></span>, would have a mass of 30 g + 2*40 g = 110 g
The new percent by mass composition of A is: <span><span><span>30g</span><span>110g</span></span>∗100%=27.3%</span>
The new percent by mass composition of B is: <span><span><span><span>80g</span><span>110g</span></span>∗100%=72.7%</span></span>
Answer:
A. The entropy of the universe is increasing.(always)
Explanation:
The Second Law of Thermodynamics states that Entropy cannot decrease, because it keeps increasing and increasing and increasing. It will always stay on the increasing side.
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72.6g
The density formula is density equal mass divided by volume (d=m/v) to solve this you must get the mass by itself. You do this by multiply volume on both sides which then gets you density times volume equal mass. Then you can plug in the numbers which is 1.20g/mL x 60.5 mL the mL cancels out which leaves you with grams and thus you have 72.6g.
Ag+ and Pb+2 are two cations that are suggested as producing insoluble halide salts when studying salts containing the halide anions, cl- and br-. First, the charge's number is provided.
Neutral binary salts, also referred to as halide salts, are mixtures of metals and non-metals. The non-metal behaves in a reduced oxidation state at all times. They are the outcome of mixing a hydroxide and hydracid. halide salts of haloids are produced by the reaction of a hydroxide and a hydracid.
Ions are cations with positive charges. They emerge when the electrons of an elemental metal are lost. However, they don't lose any protons; they only lose one or more electrons. To denote a cation, the charge is superscripted following the element name or chemical formula.
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Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
